Vaša košarica je prazna
The equation of a circle
A straight line can be given with a linear equation in two variables in the planar Cartesian coordinate system. The coordinates of the points of the straight line and only these satisfy this equation.
Our aim now is to describe another special point set of the plane, the circle unambiguously with an equation that is satisfied only by the coordinates of the points of the circle (boundary line of the circle).
As a circle is unambiguously defined in the plane by its centre and its radius, let the centre \latex{K(u; v)} and the radius \latex{r (r \gt 0)} of the circle \latex{ k } be given. The point \latex{P(x; y)} lies on the circle \latex{ k } if and only if \latex{KP = r} (Figure 42).
\latex{K(u;v)}
\latex{P(x;y)}
\latex{ x }
\latex{ 0 }
\latex{ y }
\latex{ k }
\latex{ r }
Figure 42
Based on the relation relating to the distance of two points
\latex{KP=\sqrt{(x-u)^2+(y-v)^2}=r ,}
which is equivalent to the equation
\latex{(x-u)^2+(y-v)^2=r^2}
obtained after the squaring.
We obtain that the point \latex{ P }(\latex{ x }; \latex{ y }) lies on the circle (boundary line of the circle) with the centre \latex{ K }(\latex{ u }; \latex{ v }) and with the radius \latex{ r } if and only if
\latex{(x-u)^2+(y-v)^2=r^2.}
This relation is the equation of the circle with the centre \latex{K(u; v)} and with the radius \latex{ r }.
By performing the operations of squaring the equation of the circle can also be written as follows:
By performing the operations of squaring the equation of the circle can also be written as follows:
\latex{x^2 + y^2 – 2ux – 2vy + u^2 + v^2 – r^2 = 0.}
Note: The deduction of the equation of the circle implies that for the coordinates of any point \latex{Q(x; y)} of the open disc (the boundary line is not included) with the centre \latex{K(u; v)} and only for those
\latex{(x-u)^2+(y-v)^2\lt r^2},
and for the coordinates of the points of the closed disc (the boundary line is included) and only for those
\latex{(x-u)^2+(y-v)^2\leq r^2}.
\latex{ 0}
\latex{ y }
\latex{ x }
\latex{ 4 }
\latex{r=4}
\latex{ 1}
\latex{ -1}
\latex{ -4}
\latex{ -1}
\latex{ -4}
\latex{ 1}
\latex{ 4}
Figure 43
Example 1
Let us give the equation of the circle with the origin as the centre and with a radius of \latex{ 4 } units.
Solution
Now \latex{ u = 0 }, \latex{ v = 0 } and \latex{ r = 4 } (Figure 43), thus the equation of the circle in question is
\latex{x^2 + y^2 = 16.}
Example 2
Let us set up the equation of the circle \latex{ k } with the centre \latex{K(–2; 3)} and with a radius of \latex{ 5 } units, and let us determine the coordinates of the points of the coordinate axes that lie on \latex{ k }.
Solution
\latex{ u = –2 }, \latex{ v = 3 }, \latex{ r = 5 }, thus the equation of \latex{ k } is
\latex{(x+2)^2+(y-3)^2=25}.
The circle \latex{ k } intersects the \latex{ x }-axis at the points where second coordinate is \latex{ 0 }, thus their first coordinates satisfy the equation
\latex{(x+2)^2+(-3)^2=25}.
By subtracting \latex{ 9 } from both sides and then by rearranging it so that one side is equal to \latex{ 0 }, we obtain
\latex{(x+2)^2-16=0}.
The left-hand-side of the equation can easily be factorised
\latex{(x + 2 – 4)(x + 2 + 4) = 0,}
and then after adding:
\latex{(x - 2)(x + 6) = 0.}
\latex{P_1(-6;0)}
\latex{K(-2;3)}
\latex{P_2(2;0)}
\latex{Q_2(0;3+\sqrt{21})}
\latex{Q_1(0;3-\sqrt{21})}
\latex{-6}
\latex{ y }
\latex{k}
\latex{0}
\latex{x}
\latex{2}
\latex{1}
\latex{-2}
\latex{3 }
Figure 44
From here \latex{x_1 = –6,x_2 = 2}, so the intersection points of \latex{ k } and the \latex{ x }-axis are: \latex{P_1(–6; 0), P_2(2; 0)} (Figure 44).
The first coordinate of the points of \latex{ k } lying on the \latex{ y }-axis is \latex{ 0 }, and
\latex{2^2 + (y - 3)^2 = 25}
holds for the second coordinates.
After squaring and rearranging so that one side is equal to \latex{ 0 }:
\latex{y^2 - 6y - 12 = 0.}
By solving this equation we obtain the following:
\latex{y_1=3-\sqrt{21},\;y_2=3+\sqrt{21} .}
So the points of \latex{ k } lying on the \latex{ y }-axis are:
\latex{Q_1(0; 3 –\sqrt{21} ),\;Q_2(0; 3 +\sqrt{21} ).}
Example 3
Let us give the equation of the circumscribed circle of the triangle \latex{ ABC }, if \latex{A(11; 10), B(4; 3), C(8; –5).}
Solution I
The centre of the circumscribed circle of the triangle is the intersection point of the perpendicular bisectors of the sides, therefore the centre is obtained as the intersection point of two perpendicular bisectors. The radius of the circumscribed circle is the distance of the centre and one of the vertices (Figure 45).
\latex{A(11; 10)}
\latex{B(4;3)}
\latex{C(8;-5)}
\latex{x}
\latex{ y }
\latex{ 10 }
\latex{ 3 }
\latex{ 1 }
\latex{ 0 }
\latex{ -5 }
\latex{ 1 }
\latex{ 4 }
\latex{ 8 }
\latex{f_{AB}}
\latex{ 11}
\latex{K}
\latex{r}
\latex{F_c}
\latex{F_a}
\latex{f_{BC}}
Figure 45
So the steps of the solution are:
- As \latex{\overrightarrow{AB} (-7;-7)}, a normal vector of \latex{f_{AB}} is \latex{\vec{n}_c(1;1) }, and a point of it is the midpoint of the side \latex{ AB }. Thus the equation of \latex{f_{AB}} is
\latex{ x + y = 14 }.
\latex{\overrightarrow{BC} (4;-8) }, therefore \latex{\vec{n}_a(1;-2) } is a normal vector of \latex{ f_{BC} }, and a point of it is the midpoint \latex{F_a(6; –1)} of the side \latex{ BC }.
Thus the equation of \latex{ f_{BC} } is
\latex{ x – 2y = 8 }.
- The coordinates of the centre are the solutions of the simultaneous equations below:
\latex{\begin{rcases}x+y=14\\ x-2y=8\end{rcases}}.
By subtracting the second equation from the first equation
\latex{ 3y = 6 }, which implies \latex{ y = 2 }.
If we add the second equation to twice the first equation, then
\latex{ 3x = 36 } is obtained, which implies \latex{ x = 12 }.
The centre of the circle is: \latex{ K }(\latex{ 12 }; \latex{ 2 }).
- \latex{r^2 = KA^2 = (12 – 11)^2 + (2 – 10)^2 = 1 + 64 = 65.}
- The equation of the circumscribed circle of the triangle \latex{ ABC } is
\latex{(x - 12)^2 + (y - 2)^2 = 65.}
Solution II
We performed the calculations of the previous solution based on geometric considerations. Now we will show a wholly algebraic solution.
Let the centre of the circle in question be \latex{K(u; v)}, and let its radius be\latex{ r }. Then its equation is:
\latex{(x -u)^2 + (y - v)^2 = r^2.}
The vertices of the triangle lie on the circle, which means exactly that the coordinates satisfy the equation of the circle
\latex{(11-u)^2+(10-v)^2=r^2, \;\;\;\;\;\;\;\;\;\;\;\;\tag{1}}
\latex{(4-u)^2+(3-v)^2=r^2, \;\;\;\;\;\;\;\;\;\;\;\;\tag{2}}
\latex{(8-u)^2+(-5-v)^2=r^2. \;\;\;\;\;\;\;\;\;\;\;\;\tag{3}}
\latex{(4-u)^2+(3-v)^2=r^2, \;\;\;\;\;\;\;\;\;\;\;\;\tag{2}}
\latex{(8-u)^2+(-5-v)^2=r^2. \;\;\;\;\;\;\;\;\;\;\;\;\tag{3}}
We obtain quadratic simultaneous equations for the unknowns \latex{ u }, \latex{ v } and \latex{ r }. By squaring we obtain the following:
\latex{\begin{rcases}(1)\;\;\;\;\;\;u^2+v^2-22u-20v+221=r^2,\\(2)\;\;\;\;\;\;u^2+v^2-8u-6v+25=r^2,\\(3)\;\;\;\;\;\;u^2+v^2-16u+10v+89=r^2.\end{rcases}}
By subtracting the equation (1) from the equation (2):
\latex{ 14u + 14v – 196 = 0 }, which implies \latex{ u + v = 14 }.
Now by subtracting the equation (3) from the equation (2):
\latex{ 8u – 16v – 64 = 0 }, which implies \latex{ u – 2v = 8 }.
The solution of the simultaneous equations
\latex{\begin{rcases}u+v=14\\u-2v=8\end{rcases}}
is (see solution I): \latex{ u = 12 }, \latex{ v = 2 }.
By substituting the values of u and v into one of the equations (1), (2) and (3) we obtain the following:
\latex{r^2=65(r=\sqrt{65} )}.
Example 4
Let us determine the centre and the radius of the circle if its equation is:
- \latex{x^2 + y^2 – 2x + 6y +1 = 0;}
- \latex{x^2 + y^2 + 8x – 2y +22 = 0.}
Solution (a)
Let us rearrange the equation to the form \latex{(x – u)^2 + (y – v)^2 = r^2} to determine the centre and the radius. For this let us apply the method of completing the square on the left-hand-side of the equation given:
\latex{x^2 + y^2 - 2x + 6y + 1 = (x^2 - 2x + 1) + (y^2 + 6y + 9) - 9 =}
\latex{= (x -1)^2 + (y + 3)^2 - 9 = 0.}
\latex{= (x -1)^2 + (y + 3)^2 - 9 = 0.}
So the informative form of the equation is
\latex{(x – 1)^2 + (y + 3)^2 = 32,}
and it can be seen in this form that the centre of the circle is the point \latex{ K }(\latex{ 1 }; \latex{ –3 }), and its radius is \latex{ 3 } units.
Solution (b)
Let us again rearrange the left-hand-side of the equation:
\latex{x^2 + y^2 + 8x – 2y + 22 = (x^2 + 8x + 16) + (y^2 – 2y + 1) + 5 =}
\latex{= (x + 4)^2 + (y – 1)^2 + 5 = 0,}
\latex{= (x + 4)^2 + (y – 1)^2 + 5 = 0,}
which implies
\latex{(x + 4)^2 + (y -1)^2 = -5.}
It cannot be satisfied by any real \latex{ x } or \latex{ y }, since \latex{(x + 4)^2 \geq 0, (y - 1)^2 \geq 0.} So we find that there is no circle the equation of which would be the equation in part \latex{ b }).
The solution of example \latex{ 4 } shows that it is also worth examining in general under what circumstances will a quadratic equation in two variables without the term \latex{ xy } and with equal coefficients for the quadratic terms be the equation of a circle. The generation of the equation of a circle implies that the equation of any circle can be transformed to the form
\latex{x^2 + y^2 + Ax + By + C = 0,}
therefore it is enough to examine this type of equations. By transforming the left-hand-side
\latex{x^2+y^2+Ax+By+C=\left\lgroup x+\frac{A}{2} \right\rgroup^2+\left\lgroup y+\frac{B}{2} \right\rgroup^2 -\frac{A^2}{4}-\frac{B^2}{4}+C=0,}
\latex{\left\lgroup x+\frac{A}{2} \right\rgroup^2+\left\lgroup y+\frac{B}{2} \right\rgroup^2 =\frac{A^2+B^2-4C}{4}}.
This is the equation of a circle if and only if \latex{ A^2 + B^2 – 4C\gt0 }. Then the centre of the circle is
\latex{K\left\lgroup-\frac{A}{2};-\frac{B}{2} \right\rgroup}, its radius is \latex{r=\frac{\sqrt{A^2+B^2-4C} }{2}}.
⯁ ⯁ ⯁
Now we are going to summarise and systematise the information about the equation of a circle.
DEFINITON: In the planar Cartesian coordinate system the equation of a circle is an equation in two variables that is satisfied by the coordinates of the points \latex{P(x; y)}, which lie on the circle, and is satisfied by only these.
THEOREM: A quadratic equation in two variables is the equation of a circle in the planar Cartesian coordinate system if and only if it can be transformed to the form
\latex{x^2 + y^2 + Ax + By + C = 0}
where \latex{ A }, \latex{ B }, \latex{ C } are real numbers, which satisfy the inequality \latex{ A^2 + B^2 – 4C\gt0 }. The centre of the circle generated with this equation is \latex{K\left\lgroup-\frac{A}{2};-\frac{B}{2} \right\rgroup}, its radius is \latex{r=\frac{\sqrt{A^2+B^2-4C} }{2}}.
If \latex{r = 0 (A^2 + B^2 - 4C = 0)} , then the equation is satisfied by the coordinates of a single point, the point \latex{K\left\lgroup-\frac{A}{2};-\frac{B}{2} \right\rgroup} .
⯁ ⯁ ⯁
Now we will show an example for solving a planar geometry problem with the tools of the coordinate geometry but formulated not in the language of coordinate geometry.
Example 5
Two distinct points, \latex{ A } and \latex{ B }. are given in the plane. Let us determine the set of the points \latex{ P } of the plane for which \latex{ AP : PB = 2 : 3 }.
Solution (b)
Let us align the Cartesian coordinate system in the plane so that the point \latex{ A } is at the origin, and the point \latex{ B } lies on the \latex{ x }-axis (Figure 46).
Then the points given are: \latex{A(0; 0), B(b; 0)}. The distance ratio for an arbitrary point \latex{P(x; y)} meeting the conditions is
\latex{\frac{2}{3}=\frac{AP}{PB}=\frac{\sqrt{x^2+y^2} }{\sqrt{(x-b)^2+y^2} }}.
\latex{P(x; y)}
\latex{B(b; 0)}
\latex{A(0; 0)}
\latex{ x }
\latex{ y }
\latex{ K }
Figure 46
After squaring and multiplying
\latex{4[(x - b)^2 + y^2] = 9(x^2 + y^2).}
After performing the operations and rearranging the equation so that one side is equal to \latex{ 0 }
\latex{5x^2 + 5y^2 + 8bx - 4b^2 = 0. }
By dividing both sides by \latex{ 5 }, we get
\latex{x^2+y^2+\frac{8b}{5}\times x-\frac{4b^2}{5} =0,}
which is very much like the equation of a circle. Let us rearrange it and check whether there is a circle with this equation.
By completing the square
\latex{\left\lgroup x+\frac{4b}{5} \right\rgroup^2-\frac{16b^2}{25}+y^2-\frac{20b^2}{25} =0,}
which implies
\latex{\left\lgroup x+\frac{4b}{5} \right\rgroup^2+y^2=\frac{36b^2}{25}}.
As we performed equivalent transformations, the points and only the points \latex{P(x; y)} of the point set in question satisfy the equation above.
The point set in question is obtained as a circle the centre of which is \latex{K\left\lgroup-\frac{4b}{5};0 \right\rgroup}, its radius is \latex{r=\frac{6b}{5}}.
Notes:
- With the method applied now it can also be shown in general that a circle is the set of points in the plane the distance ratio of which from two given points is a constant different from \latex{ 1 }. This circle is called the Apollonian circle corresponding to the given points and the given ratio.
- We know from earlier that if the distance ratio from the given points is \latex{ 1 }, then the point set in question is the perpendicular bisector of the line segment defined by the given points.
Exercises
- Give the equation of the circles with the centre and radius as below:
- \latex{K(0;1),r=2 };
- \latex{K(3;1),r=5 };
- \latex{K(-2;4),r=0.5 };
- \latex{K(-1;-5),r=\sqrt{3} }.
- Give the equation of the circle with the diameter \latex{ AB }, if
- \latex{A(0;0),B(-4;0)};
- \latex{A(2;3),B(-2;5)};
- \latex{A(1;-4),B(7;0)};
- \latex{A(6;-8),B(4;6)}.
- Give the coordinates of the centre and the radius of the circle if its equation is
- \latex{x^2+y^2-4y=0};
- \latex{x^2+y^2+2x-2y+1=0};
- \latex{x^2+y^2-x+\frac{3}{2}\times y-\frac{5}{2}=0};
- \latex{x^2+y^2+x+y+36=0};
- \latex{x^2+y^2=\sqrt{12} \cdot x-\sqrt{20 }\times y-1 }.
- For which real values of the parameter \latex{ p } will the equation \latex{x^2 + y^2 – 8x + 6y + p = 0} be the equation of a circle? For which values \latex{ p } will the radius of the circle be
- \latex{1};
- \latex{3};
- \latex{1.5};
- \latex{\sqrt{5}};
- \latex{\frac{3}{2}};
- \latex{11}?
- Give the coordinates of the centre and the radius of the circle with the equation \latex{x^2 + y^2 – 4x + 2y = 31.} Give the coordinates of the points of the circle lying on the coordinate axes.
- Give the equation of the circumscribed circle of the triangle \latex{ ABC }, if
- \latex{A(0;0),B(3;0),C(0;4)};
- \latex{A(0;2),B(2;0),C(5;5)};
- \latex{A(2;-3),B(2;3),C(-3;-2)}.
- The centre of a circle is \latex{ K }(\latex{ 3 }; \latex{ 3 }), and it touches the coordinate axes. Give its equation.
